ON THE STRONGLY REGULAR GRAPH OF PARAMETERS
(99, 14, 1, 2)
SUZY LOU AND MAX MURIN
Abstract. In an attempt to find a strongly regular graph of parameters (99, 14, 1, 2)
or to disprove its existence, we studied its possible substructure and constructions.
1. Introduction
Throughout the paper, the character ∼ will denote adjacency; G will denote the
graph with the parameters under question, assuming it exists; and V will denote the
vertex set of G.
Definition 1.1. A strongly regular graph with parameters (n, k, λ, µ) is a k-regular
graph on n vertices such that a pair of vertices has λ neighbors in common if they
are adjacent, and µ neighbors in common otherwise. There are many parameter sets
for which it can be proven that no strongly regular graph exists, but for many other
parameter sets, neither existence nor existence of a corresponding strongly regular
graph has been shown. One of these parameter sets is (99, 14, 1, 2).
Though these graphs are easy to define, it is not yet well understood for which
parameter sets there exist at least one corresponding graph. Because of this lack
of understanding, it is desirable to find out whether or not a parameter set such as
(99, 14, 1, 2) might correspond to a graph. In addition, it is unknown whether there is
a Moore graph, a graph with diameter k and girth 2k +1, with 57 vertices and girth 5.
If this graph exists, it would be strongly regular and would complete the classification
of Moore graphs. Though this particular problem is not related to Moore graphs,
these two facts contribute to the interest in strongly regular graphs.
In studying this graph, we created several unsuccessful attempts at construction.
We also found certain properties, such as bounds for the chromatic number and the
size of a maximal independent set, possible substructures, and possible orders of
automorphisms. In Section 2 we will first examine the ways the strongly regular
graph of parameters (9, 4, 1, 2) could potentially be a substructure. In Section 3 we
will discuss attempted constructions with Fano-planes, followed by a discussion in
Section 4 of maximal independent sets and a discussion of a triangle decomposition
in Section 5. Section 6 will contain a discussion of possible orders of automorphisms
in the graph. Section 7 will discuss the relationship of G with rotational block designs,
and finally Section 8 will discuss the structures that arise from an automorphism of
order 7.
2. The srg(9, 4, 1, 2) as a substructure
Let H be the unique strongly regular graph of parameters (9, 4, 1, 2).
Key words and phrases. Strongly regular graphs.
The project was supported by the PRIMES-USA program of MIT.
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S. LOU AND M. MURIN
Theorem 2.1. If G contains H minus an edge as a subgraph, then it contains H as
an induced subgraph.
Proof. Suppose, for the sake of contradiction, that there H minus an edge was an
induced subgraph in G.
According a lemma of Wilbrink and Brouwer[1], the following equation holds for
an induced subgraph of a strongly regular graph with parameters (n, k, λ, µ), such
that the induced subgraph has N vertices, of degree d1 , . . . , dN , and M edges:
X
N N X
N
di
j−1
(n − N ) − (kN − 2M ) + λM + µ
−M −
= x0 +
xj ,
2
2
2
i=1
j=3
where xj denotes the number of vertices outside the subgraph adjacent to exactly j
vertices in the induced subgraph.
One may verify that no vertex outside of the subgraph is adjacent to more than
3 vertices in this particular subgraph; otherwise, at least one of the parameters is
violated. Therefore, applying the lemma above, x0 + x3 =5.
The induced subgraph is illustrated in Fig. 1.
Figure 1. An illustration of the induced subgraph, which we shall
prove does not exist in G.
For convenience, the induced subgraph on the vertices labeled Xa , Xb , Ya , Yb , Za ,
Zb will be referred to as the "prism." Consider the bold edges. Each of these edges
must form a triangle with another vertex. Keeping the third and fourth parameters
in mind, we find that the two vertices that form a triangle with these edges are not
adjacent to any vertex of the triangular prism and do not coincide.
Vertices X and Z share vertex Y as a common neighbor and have one more common
neighbor. Again by examining the third and fourth parameters we find that this other
ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2)
3
common neighbor is not adjacent to any vertex of the triangular prism and does not
coincide with the vertices previously mentioned.
This results in the subgraph shown in Fig. 2:
Figure 2. A more detailed subgraph that must exist in G if the induced subgraph shown in Fig. 1 exists.
Let S be the set of the 60 vertices such that they are the neighbors of a vertex in
the prism, that are themselves not in the prism.
By the fourth parameter, X and Z are each adjacent to 2 vertices in S. Similarly,
vertex Y is adjacent to no such vertices. We find that X and Z are hence adjacent
to 9 vertices that are not in S, and for both X and Z, 2 of these 9 neighbors are
already drawn in the diagram. Similarly, Y has 10 neighbors that do not belong to
S, and two of them are drawn in the diagram.
Given this information, we can directly compute x0 : it is 5. Thus, x3 = 0.
Figure 3. All the vertices that have 2 or more neighbors in S ∩ {1, 2, 3}
Consider a vertex v belonging to this set of five vertices. By the fourth parameter,
exactly twelve of its neighbors belong to S. We also know that v shares two neighbors
with each of X, Y , Z. How is this possible? That would seem to make 18 neighbors
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S. LOU AND M. MURIN
of v, so we must be overcounting. At least 4 neighbors of v must either be adjacent
to two of X, Y , Z, or simultaneously be adjacent to one of X, Y , and Z and belong
to S. (Recall it is impossible for a vertex belonging to S to be adjacent to more than
one of X, Y , and Z.)
That is to say, each of these five are adjacent to four of the green vertices in the
subgraph shown in Fig. 3, where H − 4, H − 5, H − 6, H − 7 are the vertices that
simultaneously are adjacent to one of X, Y , and Z and belong to S.
Choosing 4 vertices from 7 is the same as not choosing 3 from the 7, and we note
that because of the fourth parameter, at least 2 members of the set {H − 1, H −
2, H − 6, H − 7} must not be chosen, and similarly, at least two members of the
set {H − 1, H − 3, H − 4, H − 5} must not be chosen. That means that H − 1 can
never be chosen. Then H − 1 has no neighbors from the vertices constituting x0 .
However, since H − 1 does not belong to S, we can count its neighbors: 2 are shown
in the previous diagram; by the fourth parameter, 8 of its neighbors belong to S. Its
remaining neighbors must be adjacent to one or more of X, Y , and Z, but not to any
vertices of the triangular prism. But it already shares two neighbors with Y , and for
each of vertex X and vertex Z it needs only 1 more common neighbor. Then H − 1
has only 12 neighbors, contradiction.
3. Labelings with Fano planes
Definition 3.1. A Fano plane is a set of seven 3-element subsets, called lines, of
{1, . . . , 7} such that every pair of lines share exactly one element. The elements are
also called points.
There are 30 distinct Fano planes, which can be grouped into two disjoint sets of 15
Fano planes, such that two Fano-planes in the same set have the following property:
the two Fano planes share exactly one 3-element set, one Fano plane can be obtained
from the other by cyclically permuting the three elements of this shared 3-element
set. In addition, if we cyclically permute three elements of a single set of a Fano
plane, the result is a Fano plane in the same set of 15 Fano planes as the initial Fano
plane.
Suppose one of these disjoint sets of 15 Fano planes is {F0 , F1 , . . . , F14 }. Suppose
that F0 , F2n−1 , and F2n share a line for n ∈ 1, . . . , 7. Then we can label G as follows:
Call a central vertex F0 , and its 14 neighbors F0 . . . F14 , such that F0 , F2n−1 , and F2n
form a triangle.
For the other 84 vertices in the graph, suppose a vertex is the common neighbor
of Fi and Fj . If Fk is the Fano-plane such that Fi , Fj , Fk all share a line, and e is
the shared line, then label the vertex as (Fk , e).
One attempt at construction was to create rules for adjacency among the vertices
(Fk , e). However, none of the rules that were tried worked. A few rules that were the
most noteworthy were the following.
1. Consider neighbors of Fx of the form (F, l). Connect two of them if the line the
Fano-plane portion of their labellings share is in F0 . This rule is equivalent to the
impossible construction with srg(9, 4, 1, 2).
2. Connect (Fx , lm ) and (Fy , ln ) if lm and ln are disjoint.
3. Consider Fx and Fy that form a triangle with F0 . Connect a neighbor of Fx of
form (F, l) and a neighbor of Fy of form (F, l) if the Fano-plane part of their label is
the same.
ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2)
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4. The following is not a rule, but rather a set of conditions for a rule.
Let e be the line Fk shares with F0 . In that case, Fk can be obtained from F0 by
cyclically permuting e. Consider the vertex e ∩ l, which we will call P . Define the
root of a vertex (Fk , l) to be the point such that P occupies the position it previously
occupied before the rotation.
The root has the following property: Consider any one of the four lines of Fk that
do not contain P , and consider two points of that line, a and b. Then a, b, and the
root do not form a line in F0 . It is the only point with this property.
Consider (Fk , l); suppose its root is P and that it is connected to Fa and Fb . Then
consider all the lines that contain P but are not included in F0 . There are 12 such
lines. For each line lm that is one of the twelve, consider the three vertices such that
lm is included in their label. Choose one of them whose root is part of l and connect
(Fk , l) to it.
The reason these conditions cannot be met is the condition that (Fk , l) shares
exactly one vertex with Fa and with Fb : there is no way to meet this condition.
4. Independent sets and 2-block designs of parameters (22, 4, 2).
Theorem 4.1. An independent set in G of size 9 cannot be maximal.
Proof. Suppose there were such a maximal independent set, I. Let xi be the number
of vertices in V \ I adjacent to exactly i vertices in I, and let yi denote the set of
vertices with i neighbors in I. For i ≥ 6, xi = 0. This is because a vertex vj with j
neighbors in I needs 2(9 − j) + j = 18 − j common neighbors with the vertices of I,
and has 14 − j neighbors in V \ I. But if vj had a neighbor with 6 or more neighbors
in I, then it would have to have at least 19 − j common neighbors with the vertices
of I. Thus, if there were a vertex with 6 or more neighbors in I, it would have no
neighbors in V \ I, which is clearly impossible.
Lemma 4.2. Related to the above observation, for all vertices vj in V \ I, if a, b, c,
and d are the number of neighbors of vj with, respectively 2, 3, 4, and 5 neighbors in
I, a + 2b + 3c + 4d = 4. This also means that a vertex in V \ I can never serve as a
common neighbor for a vertex in y5 and a vertex with 2 or more neighbors in I.
Proof. A vertex vj with j neighbors in I needs 2(9−j)+j = 18−j common neighbors
with the vertices of I. The actual number of common neighbors is 2a + 3b + 4c + 5d +
(14 − j − a − b − c − d). Setting this equal to 18 − j yields the above.
We have
x1 + x2 + x3 + x4 + x5 = 90
x1 + 2x2 + 3x3 + 4x4 + 5x5 = 9 ∗ 14 = 126
9
x2 + 3x3 + 6x4 + 10x5 = 2
= 72.
2
In addition, we have x5 ≤ 3. Suppose, on the contrary, we had 4 vertices A, B, C, D
such that each had 5 neighbors in I. If A and B shared only one common neighbor,
no vertex could have more than 4 neighbors in I without sharing 3 neighbors with
A or B. Thus, they must share 2 common neighbors. Then C and D must both be
adjacent to the vertex of I adjacent to neither A nor B, as well as 2 vertices adjacent to
A only and 2 vertices adjacent to B only. Then C and D share 3 common neighbors,
contradiction. Note that if x5 = 3, then any pair of vertices with 5 neighbors in I
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S. LOU AND M. MURIN
share two common neighbors in I. In this case, 6 vertices in I have 2 neighbors in
y5 , and 3 vertices in I have 1 neighbor in y5 .
The only solutions to the above equations are: (78, 0, 0, 12, 0), (77, 0, 6, 4, 3),
(77, 2, 0, 10, 1), (76, 4, 0, 8, 2), (77, 1, 3, 7, 2), (79, 3, 3, 5, 3), (75, 6, 0, 6, 3).
However, none of these solutions work. In the first solution, (78, 0, 0, 12, 0), consider
the 12 vertices with 4 neighbors in I. For every time a vertex in I is adjacent to such
a vertex, it gains a common neighbor with a different vertex in I 3 times. In total, it
must gain 2 ∗ 8 = 16 common neighbors in I, but 16 is not divisible by 3, so this is
impossible.
Now note that the number of edges from y2 ∪ y3 to y4 ∪ y5 is at most |y2 ∪ y3 |,
because each vertex of y2 ∪ y3 has at most one neighbor in y4 ∪ y5 . On the other
hand, it is also at least |y4 | − |y5 |, because at most |y5 | members of y4 have a neighbor
in y5 , and the rest must have a neighbor in y2 ∪ y3 . For the third, fourth, and fifth
solutions, this causes an immediate contradiction.
In the second solution, (77, 0, 6, 4, 3), Lemma 4.2 implies that the 3 vertices in y5
each have 2 neighbors in y3 , and none in y4 . Thus every vertex in y4 ∪ y5 must have
a neighbor in y2 ∪ y3 , but this means there are at least 7 edges between y2 ∪ y3 and
y4 ∪ y5 , while on the other hand there are at most 6; contradiction.
In the sixth solution, (79, 3, 3, 5, 3), consider the 5 vertices in y4 . As mentioned
before, 6 vertices in I have 2 neighbors in y5 , and 3 vertices in I have 1 neighbor in
y5 . Each vertex in y4 must have at least 2 neighbors in I that have 1 neighbor in
y5 , or else it shares at least 3 neighbors with a vertex in y5 . Thus, each serves as a
common neighbor either 1 or 3 times for the 3 vertices in I that have 1 neighbor in
y5 . These 3 vertices share a common neighbor a total of 6 times, so we see that each
of the 5 vertices in y4 has exactly 2 neighbors out of these 3 vertices in I. Then none
of them is adjacent to a vertex in y5 . Then the only way for Lemma 4.2 to be fulfilled
with respect to the vertices in y5 is for a vertex to be adjacent to 2 vertices in y3 , or
1 vertex in y3 and 2 vertices in y2 . Then 2 vertices in y5 share a common neighbor in
y3 or y2 , as well as 2 common neighbors in I, contradiction.
In the seventh solution, (75, 6, 0, 6, 3), as above, the 6 vertices in y4 must each be
adjacent to exactly 2 vertices in I that have 1 neighbor in y5 . Thus, no edges exist
between y4 and y5 . Then to fulfill the lemma, each vertex in y5 has 4 neighbors in
y2 . Then 2 vertices in y5 share at least 2 neighbors in y2 as well as 2 neighbors in I,
contradiction.
Thus, all possibilities lead to a contradiction.
Theorem 4.3. The largest independent set of a strongly regular graph of parameters
(99, 14, 1, 2) has size at most 22. If it has size 22 then every vertex not belonging to
the independent set has exactly 4 neighbors in the independent set.
Proof. Let I be a maximal independent set with n vertices. The set S of all vertices
with at least one neighbor in I has size 99 − n. The number of edges between S and
I is 14n.
Suppose P
S = {s1 , . . . , s99−n }. Let F (i) be the number of neighbors si has in I.
Therefore, 99−n
In addition, because 2 nonadjacent vertices share 2
i=1 F (i) = 14n.
P99−n
P99−n
F (i)
n
2
2
2
neighbors, i=1
=
2
=
n
−
n.
Therefore,
i=1 F (i) = 2n + 12n.
2
2
q P99−n
q
P
99−n
2
2 +12n
F (i)
14n
i=1 F (i)
By the RMS-AM inequality,
= 2n99−n
≥ i=1
= 99−n
. After
99−n
99−n
2
some algebra, this turns into −n − 5n + 594 ≥ 0, so −27 ≤ n ≤ 22. Thus, the
ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2)
7
maximum size of I is 22, as desired. Equality holds when F (i) is equal across all
values of i; thus, for all i, F (i) = 14∗22
= 4.
77
The 77 vertices outside the independent set all have 4 neighbors in S.
5. Miscellaneous.
Assuming existence, G has a unique triangle decomposition. Consider a triangle
and all triangles adjacent to it. This is illustrated in Fig. 4.
Figure 4. An illustration of a triangle in G and all adjacent triangles.
= 693 edges in the graph, there are 693
= 231 disjoint triangles
Since there are 99∗14
2
3
in the graph. We can consider a graph T on 231 vertices such that each triangle in G
is a vertex of T and two vertices in T are adjacent iff the corresponding triangles in
G share a vertex. The graph T is 18-regular, and because G has diameter 2, T has
diameter 3. In the above diagram, a central vertex in T is called v.
Lemma 5.1. The chromatic number of G is between 5 and 11.
Proof. The lower bound of 5 is a direct consequence of the fact that the maximum
size of an independent set in G is 22.
By the second parameter, there is a perfect matching between the vertices Xi and
Yi . Similarly, there is a perfect matching between the vertices Yi and Zi , and Xi and
Zi . Then these edges determine disjoint cycles of total length 36, each cycle of length
divisible by 3.
As a result, the induced subgraph on the 36 vertices is 3-regular. By Brook’s
Theorem, these vertices can be 3-colored. Assign them an arbitrary 3-coloring.
Let the set V 0 = V \ {X1 , . . . , X12 , Y1 , . . . , Y12 , Z1 , . . . , Z12 }. Now, consider the
vertices of V 0 . By the fourth parameter, each one has two neighbors of the form Xi ,
two neighbors of the form Yi , and two neighbors of the form Zi . Thus, the induced
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S. LOU AND M. MURIN
subgraph on these vertices is 8-regular. Again by Brook’s Theorem, these vertices
can be 8-colored. Assign them an arbitrary 8-coloring with colors that have not been
used before.
Now consider X, Y , and Z, which have not been assigned colors. Assign them 3
distinct colors from the 8-coloring on the vertices in V 0 .
Thus, the graph is 11-colorable.
Let us shift our attention again to T , so that a vertex refers to a triangle, and
a G-vertex refers to a vertex in the traditional sense. Consider the set of vertices
distance 2 from v. Now define the following:
α : The number of such vertices connected to exactly one G-vertex distance 1 from
v.
α − vertex : A vertex with the above property.
β : The number of such vertices connected to exactly two G-vertices distance 1
from v.
β − vertex : A vertex with the above property.
γ : The number of such vertices connected to exactly three G-vertices distance 1
from v.
γ − vertex : A vertex with the above property.
One may easily verify the following equations:
α + β = 180
β + 3γ = 36
α − 3γ = 144.
Also, the number of vertices distance 3 from v is 32 − γ = 20 + β3 . The value γ is an
integer between 0 and 12. Note that γ cannot be 11: as we noted before, the perfect
matchings between the vertices of form Xi , Yi ; Yi , Zi , and Xi , Zi fall into disjoint
cycles of lengths divisible by 3 and summing to 36. But if γ = 11, we have 11 3-cycles
and three loose edges that do not fall into cycles; contradiction.
If γ = 12 for all vertices in T , then this reduces to the impossible labeling with
srg(9, 4, 1, 2). This is apparent after relabeling the previous diagram as in Fig. 5.
Figure 5. A relabeling of the diagram in Fig. 4 that demonstrates
the connection between γ = 12 and the first structure discussed.
ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2)
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6. Possible orders of automorphisms of G
Theorem 6.1. srg(99, 14, 1, 2) has no automorphisms of p > 14, where p is prime.
Proof. An automorphism of the graph G of order p must have at least one orbit of
order p, since p is prime. However, not every point of the graph can be in such an
orbit, since p - 99. Since the graph is connected, at least one point in an orbit must
connect to a point P not in an orbit. However, by applying the automorphism, we
can see that the P connects to every point in the orbit, so deg P = 14 ≥ p > 14.
This is a contradiction, so no automorphisms of order p exist.
Theorem 6.2. srg(99, 14, 1, 2) has no automorphisms of order 13.
Proof. Assume such an automorphism π exists. Then, as before, we must have at
least one automorphism of order 13, and points not in any orbits. By connectedness,
there must exist a point P which connects to an orbit; thus, P must connect to every
point in that orbit. P must also connect to exactly one more point, which cannot
be in an orbit. By the property λ = 1, P must share a common neighbor with
each of these points. Thus, two points A and B in the orbit must be connected to
each other. Since π has order 13, and A and B are in the same orbit, there exists a
positive integer n < 13 such that π n (A) = B. Since π is an automorphism, π n is also
an automorphism. Thus, B connects to π n (B). However, P and B are connected,
and have two common neighbors: A and π n (B). This is a contradiction, so π cannot
exist.
Theorem 6.3. srg(99, 14, 1, 2) has no automorphism of order 11.
Proof. Assume that some such automorphism π exists. Let n be the number of orbits
of size 11 of π. Then, 1 ≤ n ≤ 9. First, examine the case that n < 9. In this case,
orbits of size 1 exist. By connectedness, there must be a point P that connects to an
orbit. Each of the points in the orbit must connect to another neighbor of P . Since
P has 3 neighbors outside of the orbit and 11 inside it, there must be two points in
the orbit that connect. As before, contradiction. Thus, n = 9.
Let us label the orbits A1 through A9 . Then, let us define a matrix M by Mij being
the number of points in Ai that any point in Aj connects to. Note that Mij = Mji ,
so M is symmetric.
If any two points in the same orbit Ai , P and Q, are connected, then there exists
a j such that P = π j (Q). Then, since π is an automorphism, π j (P ) ∼ π j (Q) = P .
π j (P ) is also in Ai . If the same process is repeated with P and π j (P ), we get back
Q. Therefore, for any point in Ai , all of its neighbors in Ai can be paired, and thus
Mii is even for all i.
Let us consider the eigenvalues of M . The eigenvalues of the adjacency matrix
of G are 14, 3, and −4. Any eigenvector of M corresponds to an eigenvector of
G, the correspondence being to set every point in Ai to the corresponding value in
the eigenvector of M . By the definition of M , the eigenvalue must also be equal:
therefore, the eigenvalues of M must be in the set {14, 3, −4}. Since the adjacency
matrix of G has the eigenvalue 14 with multiplicity 1, M can have this eigenvalue
with multiplicity at most 1; the vector h1, 1, 1, 1, 1, 1, 1, 1, 1i has this eigenvalue, so
the multiplicity of the eigenvalue 14 is exactly 1. Next, note that the sum of the
eigenvalues is equal to the trace, and that the diagonal of M must contain only
positive even integers. Thus, the sum of the eigenvalues must be an even positive
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S. LOU AND M. MURIN
integer that is a sum of 14 and eight values from the set {3, −4}. The only such even
positive sums are 38, 24, and 10.
Let us now consider the matrix M 2 . Since Mij counts the number of ways to get
from one specific point in Ai to any point in Aj by a path of length exactly 1, (M 2 )ij
counts the number of ways to get from any specific point in Ai to some point in Aj
along a path of length exactly 2. In other words, (M 2 )ij is the number of common
neighbors one specific point of Ai has with all of the points of Aj . First, consider
the case that i = j. Any point has degree 14, so it has 14 common neighbors with
itself. Then, any point of the orbit Ai is connected to Mii points on the same orbit
by the definition of M . For each one it is connected to, it has 1 common neighbor;
otherwise, it has 2. Thus, (M 2 )ii = 14 + Mii + 2(10 − Mii ) = 34 − Mii . If i 6= j, then
(M 2 )ij = Mij + 2(11 − Mij ) = 22 − Mij , similarly. Therefore, (M 2 + M )ij = 34 if
i = j and 22 otherwise.
By definition,
2
(M )ii =
9
X
Mij2 = 34 − Mii .
j=1
This implies that every value in M must be at most 5.
Since the degree of any vertex is 14, the sum of the values of any row must be equal
to 14. By trying every possibility, it can be shown that there are only seven possible
rows of M that fulfill these two equations, up to permutation:
0, 4, 3, 2, 1, 1, 1, 1, 1
0, 4, 2, 2, 2, 2, 1, 1, 0
0, 3, 3, 3, 2, 1, 1, 1, 0
0, 3, 3, 2, 2, 2, 2, 0, 0
2, 4, 2, 2, 1, 1, 1, 1, 0
2, 3, 3, 2, 2, 1, 1, 0, 0
4, 2, 2, 1, 1, 1, 1, 1, 1
The first value in each listing must lie on the diagonal.
Note that the value 5 occurs in no row. Therefore, it is never possible to have a
value of 5 in M . Thus, the sum of the eigenvalues can never be 38, so it must be
either 24 or 10.
Since we know every possible row, we can now try every possibility. No such matrix
exists. Therefore, there is no automorphism of order 11.
7. Block designs of the parameters (22, 4, 2).
Definition 7.1. A block design of parameters (22, 4, 2) is comprised of a set S of 22
values (for convenience let them be the integers from 0 to 21), called treatments, and
a set B of 77 4-subsets of S, called blocks, such that every value k in S is in exactly
14 members of B; and every pair of distinct values in S is in exactly 2 members of
B.
Let G be some srg(99, 14, 1, 2) that has an independent set S of size 22. Then, let
G\S be B. Let the graph G0 be G with every edge between two points of B removed;
G0 is bipartite with parts S and B. As noted earlier, every vertex in B must have
degree 4 in G0 . Let B 0 = {{k ∈ S | k ∼ b} | b ∈ B}. Since the srg parameter µ is
2, every pair of members of S must have two common neighbors. Thus, (S, B 0 ) is a
ON THE STRONGLY REGULAR GRAPH OF PARAMETERS (99, 14, 1, 2)
11
(22, 4, 2)-block design. Note, however, that not every block design corresponds with
a potential G0 : some block designs have repeated blocks, or blocks that share three
elements: this would lead to two points in B having four or three common neighbors,
respectively.
Every possible graph that has an independent set of size 22 can thus be associated
with a block design. One notable family of block designs is the family of cyclic, or
1-rotational, block designs, which have the property that if any block {a1 , a2 , a3 , a4 }
is in the design, then the block {a1 + 1, a2 + 1, a3 + 1, a4 + 1} is also in the design
(addition modulo 22). Some members of this family have repeated blocks or blocks
that share three members; these do not form viable graphs.
Every block in a cyclic block design is part of a family of blocks produced by adding
one to each element. For a block b, let f (b) be the block generated by adding 1 to
every element of b modulo 22. Clearly, f 22 (b) = b. Thus, the order of b under f must
divide 22. If the order was 1, then for some k in b, k + 1, k + 2, k + 3, and k + 4
would also have to be in b, contradiction. A similar contradiction arises for an order
of 2. Thus, the orders of all blocks must be either 11 or 22. If the order is 11, then
the block must be equal to {k1 , k1 + 11, k2 , k2 + 11}; if it is not of this form, it must
have order 22.
Another family of block designs is the family of 2-rotational block designs, with
the property that if any block {a1 , a2 , a3 , a4 } is in the design, then the block {a1 +
2, a2 + 2, a3 + 2, a4 + 2} is also in the design. Thus, the family of 2-rotational block
designs is a superset of the family of cyclic block designs. Defining the function g(b)
as f (f (b)), with f as before, then every block in such a design must have g 11 (b) = 1.
As before, no block can have an order of 1, so every block in such a block design has
order 11.
An attractive potential construction of G from a 2-rotational block design is to
require that if two points in G0 labeled by blocks b1 and b2 are connected, then so
are g(b1 ) and g(b2 ). This, however, would mean that g would be an automorphism
of G with order 11. As shown above, this is impossible, so the most attractive
construction does not work. It might still be possible to create an srg(99, 14, 1, 2)
from a 2-rotational block design through some method.
8. Automorphisms of order 7.
Assume that some automorphism π of G of order 7 existed. Then, since 7 does not
divide 99, there must be at least one orbit of size 1. Since the graph is connected,
at least one orbit of size 1 must connect to an orbit of size 7. Let us call the point
in this orbit P , and let us call the orbit of size 7 to which P connects A. Since P
is connected to a point in A, it must have exactly one common neighbor with that
point. No element of A can connect to any other orbits of size 1, because then P
would have seven common neighbors with that orbit. Thus, the common neighbor
must be in an orbit of size 7; call that orbit B. It is connected to A and P .
Therefore, P connects to two orbits of size 7, A and B; P has degree 14, so it does
not connect to any other orbits of size 1. Thus, any other orbits of size 1 must be
connected to two orbits of size 7 similarly. Thus, if there are k orbits of size 1, there
must be at least 2k orbits of size 7. This requires 15k points, so k ≤ 99/15 < 7. The
number of orbits of size 7 is (99 − k)/7, which must be an integer, so k must be 1
modulo 7. Therefore, k must equal 1, and thus P is the only orbit of size 1.
12
S. LOU AND M. MURIN
Let us label the members of A and B as A1 through A7 and B1 through B7 respectively, so that π(Ai ) = Ai+1 . We know that every point in A must connect to a point
in B, so WLOG let Ai connect to Bi . No Ai can also connect to any Bj for i 6= j,
since that would cause Ai and P to have two common neighbors, even though they
are connected to each other. Similarly, Ai is not connected to Aj for any j.
Since Ai and Bj are not connected for i 6= j, they must share two common neighbors, one of which must be P . Let this common neighbor be called Qi,j . Note
that Ai+1 and Bj+1 have the common neighbor Qi+1,j+1 , but also Ai+1 = π(Ai ) and
Bj+1 = π(Bj ), so Qi+1,j+1 = π(Qi,j ).
α
= Q−α+β+k,α+β+k =
Let us define Cβα as Q−α+β,α+β for α ∈ {1, 2, 3}. Then, Cβ+k
k
k
α
α
α
π (Q−α+β,α+β ) = π (Cβ ). Also, Ai ∼ Ci+α and Ci ∼ Bi+α . Define Dβα as Qα+β,−α+β
α
α
for α ∈ {1, 2, 3}. As before, Dβ+k
= π k (Dβα ). Also, Bi ∼ Ci+α
and Ciα ∼ Ai+α .
Now, note that any Ai and Aj for unequal i and j must share a common neighbor;
0
. As
let this be Ri,j = Rj,i . Let us similarly define the neighbor of Bi and Bj as Ri,j
k
α
before, Ri+k,j+k = π (Ri,j ). Then, we can define Eβ as R−α+β,α+β for α ∈ {1, 2, 3}.
0
Therefore, as before, E(α β + k) = π k (Eβα ). Similarly define Fβα as R−α+β,α+β
. Once
α
k
α
again, F( β + k) = π (Fβ ).
Let us now define a 15 by 15 matrix M . The columns and rows correspond to P ,
A, B, C 1 , C 2 , C 3 , D1 , D2 , D3 , E 1 , E 2 , E 3 , F 1 , F 2 , and F 3 in that order, and Mij is
defined as the number of times any point in the ith orbit is adjacent to some point
in the jth orbit.
Lemma 8.1. Either Mii = 0 or Mii = 2.
Proof. If there is an edge within Ci , then clearly there is a 7-cycle within Ci . If
there are 2 7-cycles, then it is easy to verify that the parameters of the graph are
violated.
Further note that no point in P , A, or B is adjacent to any other point in its own
orbit, so M11 = M22 = M33 = 0. Thus, the trace is at most 12 × 2 = 24, and as
before, must be even. The eigenvalues of M must be a subset of the eigenvalues of
G, as shown in the proof of Theorem 6.3, so M has one eigenvalue equal to 14, the
rest being either 3 or −4. If a of the eigenvalues are −4, then the trace is equal to
14 + 42 − 7a, which must be divisible by 7. Thus, the trace is either 0 or 14.
9. Acknowledgements
We are grateful to MIT PRIMES-USA for providing us with this project, and to
Dr. Peter Csikvari for suggesting the problem and mentoring us.
References
[1] H. A. Wilbrink and A. E. Brouwer, A (57, 14, 1) strongly regular graph does not exist, Indagationes Mathematicae (Proceedings) 86 (1983), 117–21.